STRUCTURE OF Co-59 AND Co-56
By Prof.Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. STRUCTURE OF STABLE Co-59 WITH S = -7/2 After a detailed analysis of the structure of Cr-48 one concludes that the structures of Co-59 and Co-56 have the core which is similar to the structure of Cr-48 with S=0. (See my STRUCTURE OF Cr-52 AND Cr-48 ). Although the isolated Cr-48 has an unstable structure, here the high symmetry of Cr-48 plays the role of the very stable core for the structure of Co-59. In the diagram of Co-59 you see the core of Cr-48 with S=0 in which we add the additional deuterons like the p25n25 of S=-1 the p26n26 of S=-1 and the p27n27 of S= -1. Note that this deuteron is shown in the second diagram of the sixth horizontal plane . Thus the total spin of such additional deuterons is S = -3 In order to get the stable structure of Co-59 with S= -7/2 we add the five extra neutrons like the n28(+1/2), the n29(+1/2), the n30(-1/2), the n31(-1/2), and the n32(-1/2). Under this condition we get S = -3 +1/2 +1/2 -1/2 – 1/2 -1/2 -1/2 = - 7/2 The n30 and n32 are not shown because they are in front of p14 and behind the p16 respectively . Note that the n31 is shown in the second diagram of the simple sixth horizontal plane. ' DIAGRAM OF Co-59 WITH S = -7/2' n25……….p12..........n12……….p26 p25……..n11..........p11 ……….n26 Sixth horizontal plane of 8 nucleons p22..........n10..........p10..........n24 n22..........p9............n9..........p24 Fifth horizontal plane of 8 nucleons n14..........p8............n8............p16 p14..........n7...........p7...........n16 Fourth horizontal plane of 8 nucleons p13.............n6...........p6............n15 n13..........p5...........n5............p15 Third horizontal plane of 8 nucleons n21……….p4...........n4……….p23 p21………n3..........p3 ……… n23 Second horizontal plane of 8 nucleons n2...........p2……….n29 n28……….p1...........n1 First horizontal plane of 6 nucleons ' ' Second diagram of the simple sixth horizontal plane at which all 11 nucleons are shown n25……….p12……….n12………p26 p25……….n11……….p11……….n26 n31………p27………..n27 STRUCTURE OF THE UNSTABLE Co-56 WITH S = +4 Here the extra neutrons are not five as in the stable Co-59 but only two extra neutrons of positive spins. Note that the deuteron n27p27 with s= +1 is not shown because it is in front of p1n1. It is of interest to note that the deuterons n25p25 with S=+1 , n26p26 with S=+1 and the n27p27 with S=+1 of the first horizontal plane of positive spins contribute to the total spin S=+4 because the two extra neutrons have positive spins . That is S = +1 +1 +1 +2(+1/2) = +4 ' ' How the unstable Co-56 of S = +4 decays to Fe-56 of S =0 ''' In the absence of the three extra nucleons the pp repulsions overcome the pn bonds and the Co-56 decays to the stable structure of Fe-56. (See my STRUCTURE OF Fe-56, Fe-55 AND Mn-55 ). So one proton decays to one neutron of the stable Fe-56. '''DIAGRAM OF Co-56 WITH S = +4 (The nucleons p19, n19, n20, and p20 of the alpha particles with S=0 are not shown here because they are behind the n6, p6, p8, and n8 respectively. Also the n17, p17, p18, and n18 are not shown because they are in front of p5, n5, n7, and p7 respectively. It is of interest to notice that the deuteron p27n27 with S =+1 is not shown because it is in front of p1n1. Also the extra n28(+1/2) and n29(+1/2) are not shown because they are in front of p26 and p15 respectively p12..........n12 n11..........p11 Sixth horizontal plane p22.........n10..........p10..........n24 n22..........p9............n9..........p24 Fifth horizontal plane n14..........p8............n8............p16 p14..........n7............p7...........n16 Fourth horizontal plane p13...........n6...........p6............n15 n13..........p5...........n5............p15 Third horizontal plane n21……….p4............n4……….p23 p21………n3...........p3 ……… n23 Second horizontal plane p25...........n2............p2……….n26 n25……….p1...........n1………p26 First horizontal plane Category:Fundamental physics concepts